We first claim that the family \((\tau _z\otimes \mathrm {id})(u_z)\) , \(z \in \mathfrak {Z}_n\) , is a linear basis in \(C(G_n)\) . This follows from that \(z_1 \ne z_2\)  implies \(h((\tau _{z_1}\otimes \mathrm {id})(u_{z_1})^* (\tau _{z_2}\otimes \mathrm {id})(u_{z_2})) = 0\)  with a unique Haar state \(h\)  thanks to [1]} together with (REF ).
Assume that \(\chi _1(u) = \chi _2(u)\)  in \(C(G)\)  with \(\chi _1,\chi _2 \in K_{-1}^\mathrm {lin}(\vartheta ^t)\) . By definition \(\chi _1(u_n) = \chi _2(u_n)\)  in \(C(G_n)\)  for every \(n \ge 0\) . It follows, from the above claim, that \(\chi _1(z) = \chi _2(z)\)  holds for every \(z \in \mathfrak {Z} = \bigsqcup _{n\ge 0}\mathfrak {Z}_n\) . Thanks to [2]} (or Theorem REF  of this paper) we conclude that \(\chi _1 = \chi _2\) .
Assume that \(\chi _i \rightarrow \chi \)  in \(K_{-1}^\mathrm {ln}(\vartheta ^t)\)  with respect to the weak\(^*\) -topology (n.b., the weak\(^*\) -topology is metrizable; see [2]}). Let \(n\ge 0\)  be arbitrarily fixed and choose an increasing sequence \(\mathfrak {F}_k\)  of finite subsets of \(\mathfrak {Z}_n\) . We have
\(\Vert \chi _i(u_n) - \chi (u_n)\Vert _{C(G_n)}&\leqq \sum _{z \in \mathfrak {Z}_n} |\chi _i(z)-\chi (z)| \\&\leqq 2\sum _{z\in \mathfrak {F}_k} |\chi _i(z)-\chi (z)| + 2\Big (1 - \sum _{z\in \mathfrak {F}_k} \chi (z)\Big ),\) 

and hence
\(\varlimsup _{i\rightarrow \infty } \Vert \chi _i(u_n) - \chi (u_n)\Vert _{C(G_n)} \le 2\Big (1 - \sum _{z\in \mathfrak {F}_k} \chi (z)\Big ) \rightarrow 0 \quad (\text{as $k \rightarrow \infty $}).\) 
Therefore, we conclude that \(\Vert \chi _i(u_n) - \chi (u_n)\Vert _{C(G_n)} \rightarrow 0\)  as \(i\rightarrow \infty \)  for every \(n\ge 0\) . This means that \(\chi _i(u) \rightarrow \chi (u)\)  in \(C(G)\)  as \(i\rightarrow \infty \) . Hence \(\chi \mapsto \chi (u)\)  is continuous.
Assume that \(\chi _i(u) \rightarrow \chi (u)\)  as \(i\rightarrow \infty \) , that is, \(\Vert \chi _i(u_n)-\chi (u_n)\Vert \rightarrow 0\)  as \(i\rightarrow \infty \)  for all \(n \ge 0\) . Since \(v_z := (\tau _z\otimes \mathrm {id})(u_z) \in \mathbb {C}[G_n]\)  and \(h\)  is faithful on \(\mathbb {C}[G_n]\)  (see [1]}), we see that \(h(v_z^* v_z) > 0\)  thanks to (REF ). By [1]} again we have
\(|\chi _i(z)-\chi (z)|\,h(v_z^* v_z) =\big |(\chi _i(u_n)-\chi (u_n)\,|\,v_z)_h\big | \le \Vert \chi _i(u_n)-\chi (u_n)\Vert \,h(v_z^* v_z)^{1/2},\) 
where \((a\,|\,b)_h := h(b^* a)\)  for \(a,b \in \mathbb {C}[G_n]\) . Hence \(\chi _i(z) \rightarrow \chi (z)\)  as \(i\rightarrow \infty \)  for all \(z \in \mathfrak {Z} = \bigcup _{n\ge 0}\mathfrak {Z}_n\) . Consequently, \(\chi _i \rightarrow \chi \)  in \(K_{-1}^\mathrm {ln}(\vartheta ^t)\)  thanks to [2]}. Hence we have shown that \(\chi \mapsto \chi (u)\)  is open.
